I hope to add more after the question is clarified a little, but here is a start:
If you have a vector space $V$, there will often be lots of subspaces.
For instance, if $v\in V$ is any nonzero element, then $\langle v \rangle=\{\alpha v\mid f \in \mathbb{F}\}$ will be a subspace of $V$, since it's clearly closed under addition and scalar multiplication. It's a one dimensional subspace, so any nonzero vector will generate this little space. If your field is of characteristic $0$, then it will also contain $2v,3v,4v,\dots nv$ because $2,3,\dots n$ are elements of the field $\mathbb{F}$. However there are many more scalars than the integers...
If you pick $w\notin \langle v\rangle$, then $\langle w \rangle$ will produce another subspace of its own, but it won't share any elements with $\langle v \rangle$ except for $0$.
If you take the $v$ and $w$ we have chosen, you can also find another subspace $\langle v,w \rangle=\{\alpha v +\beta w\mid \alpha,\beta\in \mathbb{F}\}$, which is a two dimensional subspace.
It also may be possible to find another $z\in V$ such that $\langle v,z\rangle$, but it is not equal to $\langle v,w\rangle$.
Hopefully you can see how this works for even larger collections of vectors.
If this topic is very new to you, I would highly recommend getting your head around the concept of linear independence as early as possible.
Best Answer
Probably it depends on your definition of vector space (i.e.: do you consider $\emptyset$ to be a vector space?) In my opinion, $\emptyset$ is should not be considered a vector space for various reasons, e.g. the fact that $\operatorname{span}\emptyset=\{0\}$, and thus point (1) should be included in the axioms for vector subspaces.